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mafiozo [28]
4 years ago
11

Albert's hourly pay is $8.10. He has an hourly pay of $12.15 when he works more than 40 hours. He worked 45 hours last week. Wha

t is Albert's gross pay for this pay period?
Mathematics
1 answer:
Sedaia [141]4 years ago
3 0

Answer:

$384.75

Step-by-step explanation:

Albert's hourly pay is $8.10

When he works for more than 40 hours, his hourly pay will increase to $12.15.

Last week, Albert worked for 45 hours, his hourly pay would be:

First 40 hours = $8.10 * 40 = $324

Remaining 5 hours =

$12.15 * 5 = $60.75

Albert gross pay for this period will be calculated as:

$324 + $60.75 = $384.75

Therefore, Albert gross pay for this period = $384.75

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tis a little of plain differentiation.

we know the radius of the cone is decreasing at 10 mtr/mins, or namely dr/dt = -10, decreasing, meaning is negative.

we know the volume is decreasing at a rate of 1346 mtr/mins or namely dV/dt = -1346, also negative.

so, when h = 9 and V = 307, what is dh/dt in essence.

we'll be needing the "r" value at that instant, so let's get it

V=\cfrac{1}{3}\pi r^2 h\implies 307=\cfrac{\pi }{3}r^2(9)\implies \sqrt{\cfrac{307}{3\pi }}=r

now let's get the derivative of the volume of the cone

V=\cfrac{1}{3}\pi r^2 h\implies \cfrac{dV}{dt}=\cfrac{\pi }{3}\stackrel{product~rule}{ \left[ \underset{chain~rule}{2r\cdot \cfrac{dr}{dt}}\cdot h+r^2\cdot \cfrac{dh}{dt} \right]} \\\\\\ -1346=\cfrac{\pi }{3}\left[2\sqrt{\cfrac{307}{3\pi }}(-10)(9)~~+ ~~ \cfrac{307}{3\pi } \cdot \cfrac{dh}{dt}\right]

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